Find the solution of Legendre's equation $$ (1-x^2)y'' -2xy' + l(l+1)y = 0 $$ in the form of inverse power series: $$y = \sum_{n=0}^{\infty} a_n x^{\sigma - n} $$
Hint: $\sigma = l$ and $\sigma = -(l+1)$ are the solutions
I have tried substituting the series into the equation according to the Frobenius method, but it doesn't yield the right values for sigma.
(I essentially end up with a series in $x^{\sigma - n}$ and one in $x^{\sigma - n -2}$, but due to the inverse nature of the series, after rewriting the two sums as the same power, the series previously in $x^{\sigma - n -2}$ is the one starting at a lower n. However, this series has no dependence on l, so the correct values of sigma can never be achieved by choosing n=0.)
\begin{eqnarray*} y&=& a_0 x^{\sigma}+ a_1 x^{\sigma-1}+ a_2 x^{\sigma-2}+ \cdots \\ y'&=& \sigma a_0 x^{\sigma-1}+ (\sigma -1) a_1 x^{\sigma-2}+ (\sigma -2)a_2 x^{\sigma-3}+ \cdots \\ y''&=& \sigma (\sigma -1) a_0 x^{\sigma-2}+ (\sigma -1)(\sigma -2) a_1 x^{\sigma-3}+ (\sigma -2)(\sigma -3)a_2 x^{\sigma-4}+ \cdots \\ \end{eqnarray*} Substitute these into $(1-x^2)y'' -2xy' + l(l+1)y = 0$ \begin{eqnarray*} &a_0 x^{\sigma}( \color{blue}{ -\sigma (\sigma -1)-2 \sigma + l(l+1)} ) \\ &+a_1 x^{\sigma-1}( (\sigma-1) (\sigma -2)-2 (\sigma-1) + l(l+1) ) \\ &+ x^{\sigma-2}( a_2((\sigma-2) (\sigma -3)-2 (\sigma-1) + l(l+1))+ a_0 \sigma (\sigma -1) ) \cdots =0 \end{eqnarray*} So we have $\color{blue}{ -\sigma (\sigma -1)-2 \sigma + l(l+1)}=0$.